Iterate through char array and print chars

c string ansi
ojhawkins picture ojhawkins · Sep 28, 2013 · Viewed 26.1k times · Source

I am trying to print each char in a variable.

I can print the ANSI char number by changing to this printf("Value: %d\n", d[i]); but am failing to actually print the string character itself.

What I am doing wrong here?

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int main(int argc, char *argv[])
{
  int len = strlen(argv[1]);
  char *d = malloc (strlen(argv[1])+1);
  strcpy(d,argv[1]);

  int i;
  for(i=0;i<len;i++){
    printf("Value: %s\n", (char)d[i]);
} 
    return 0;
}

Answer

mvp picture mvp · Sep 28, 2013

You should use %c format to print characters in C. You are using %s, which requires to use pointer to the string, but in your case you are providing integer instead of pointer.

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